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Let $\chi_{q}$ be a primitive Dirichlet character with modulus $q$ (see definition at wikipedia ). For example for $q=5$ we have

\begin{equation} \begin{aligned} \chi_{5,1}&=(1, 1, 1, 1, 0),\\ \chi_{5,2}&=(1, i, -i, -1, 0),\qquad\qquad \text{(1)}\\ \chi_{5,3}&=(1, -1, -1, 1, 0),\\ \chi_{5,4}&=(1, -i, i, -1, 0),\\ \end{aligned} \end{equation}

We construct a sum involving Dirichlet character $\chi_{q}$ as

\begin{equation} \begin{aligned} Q(\chi_{q})&=\sum_{k=1}^{q-1}k\chi_q(k)\qquad\qquad \text{(2)} \end{aligned} \end{equation}

Proposition A: For a complex $\chi_q$, like $\chi_{5,2},\chi_{5,4}$, \begin{equation} \begin{aligned} &\color{red}{\text{sign}(\mathrm{Re}\chi_{q}(-1))}Q(\mathrm{Re}\chi_{q})>0,\qquad\text{(3)}\\ &\color{red}{\text{sign}(\mathrm{Im}\chi_{q}(-2))}Q(\mathrm{Im}\chi_{q})>0.\qquad\text{(4)}\\ \end{aligned} \end{equation}

We are seeking a proof or a reference of the proof for this proposition.

This problem is related to a partial answer of another problem titled sign unchanged for Dirichlet polynomials? that we posted earlier at math.stackexchange.com. The motivation of studying this problem is also mentioned there.

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  • $\begingroup$ Thanks!. I changed the title as you suggested! $\endgroup$
    – mike
    Aug 7, 2019 at 6:00
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    $\begingroup$ Sure. Yes. $Q(\mathrm{Re}\chi_q)=\mathrm{Re}(Q(\chi_q))$. $\endgroup$
    – mike
    Aug 7, 2019 at 6:18
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    $\begingroup$ Assertions (3) and (4) are both false. There is a character modulo 7 for which $Q(\chi_q) = 0$. Even if you weaken the assertion a non-strict inequality $\ge 0$ rather than $> 0$, you can find characters of modulus $\le 20$ for which (3) and (4) are negative. Did you do any checking at all before you posted this conjecture here? $\endgroup$ Aug 7, 2019 at 8:31
  • $\begingroup$ @DavidLoeffler: You beat me by 58 seconds! $\endgroup$
    – GH from MO
    Aug 7, 2019 at 8:33
  • $\begingroup$ @DavidLoeffler: Thanks a lot for the comment. I am sorry that I have not done enough testing. $\endgroup$
    – mike
    Aug 7, 2019 at 16:06

1 Answer 1

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(3) and (4) are false in general, even if we weaken $>$ to $\geq$. Let $\zeta:=e^{i\pi/8}$ be a primitive $16$-th root of unity, and let $\chi$ be the unique primitive Dirichlet character modulo $17$ satisfying $\chi(3)=\zeta^5$. Then $\chi(-1)=-1$, and $$(\chi(1),\chi(2),\chi(3),\chi(4),\chi(5),\chi(6),\chi(7),\chi(8))=(1,\zeta^6,\zeta^5,\zeta^{12},\zeta^9,\zeta^{11},\zeta^7,\zeta^2).$$ However, \begin{align*}Q(\chi)&=\sum_{k=1}^{16} k\chi(k)=\sum_{k=1}^{8} (2k-17)\chi(k)\\[6pt] &=-(15+13\zeta^6+11\zeta^5+9\zeta^{12}+7\zeta^9+5\zeta^{11}+3\zeta^7+\zeta^2)\\[10pt] &\approx \ 8.84701161719 - 4.91203840222\,i. \end{align*} So $Q(\chi)$ has positive real part, even though $\chi(-1)=-1$. This contradicts (3). Similarly, if we change $\zeta^5$ to $\zeta^3$ in the definition of $\chi$, we get a counterexample to (4).

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  • $\begingroup$ Thanks a lot for the detailed answer! I noticed that you used $\chi(k)=-\chi(17-k)$. What is this (anti)symmetry? Best $\endgroup$
    – mike
    Aug 7, 2019 at 15:17
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    $\begingroup$ @mike: $\chi(17-k)=\chi(-k)=\chi(-1)\chi(k)=-\chi(k)$. $\endgroup$
    – GH from MO
    Aug 7, 2019 at 16:10
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    $\begingroup$ Got it! Thanks again! $\endgroup$
    – mike
    Aug 7, 2019 at 16:17

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